Problem: Solve for $x$, ignoring any extraneous solutions: $\dfrac{x^2 + 2x}{x + 5} = \dfrac{-2x + 5}{x + 5}$
Multiply both sides by $x + 5$ $ \dfrac{x^2 + 2x}{x + 5} (x + 5) = \dfrac{-2x + 5}{x + 5} (x + 5)$ $ x^2 + 2x = -2x + 5$ Subtract $-2x + 5$ from both sides: $ x^2 + 2x - (-2x + 5) = -2x + 5 - (-2x + 5)$ $ x^2 + 2x + 2x - 5 = 0$ $ x^2 + 4x - 5 = 0$ Factor the expression: $ (x + 5)(x - 1) = 0$ Therefore $x = -5$ or $x = 1$ However, the original expression is undefined when $x = -5$. Therefore, the only solution is $x = 1$.